Distributional second-order derivatives of $frac{e^{-|x|}}{4pi |x|}$ to show the solution of $u -Delta u=f$...
In Brezis's book "Functional Anlaysis" it is proven that the solutions of the Helmotz equation $u - Delta u=f$ where $f in L^2 (mathbb{Omega})$ belong to $H^2 (mathbb{Omega}) cap H^1 _0 (Omega)$, when $Omega$ is a bounded domain of $mathbb{R}^n$ with $partial Omega$ regular enough. I think he also takes the Dirichlet boundary conditions equal to 0, but the problem I am dealing with takes into account $Omega = mathbb{R}^n$, so it shouldn't be a problem to integrate by parts using sufficently large balls. The thing is, he proves the existence of such a solution by shifting the problem to a minum of a particular functional. He then proves this minumum (that is a weak solution of the equation) belongs to $H^{1}_0 (Omega)$. Then, later on in the book, he proves that under the hypothesis of some regularity for $partial Omega$, the solution is in $H^2(Omega)cap H^1 _0 (Omega)$, and he does that using the so called "method of Nirmberg's translations".
The fact is: I am dealing with the case $Omega= mathbb{R}^3$. I think the same results that were true for regular bounded domains should also be true in this case, I think it should be enough to consider the limit of domains $Omega_N = B(0,N)$ where the proof of Brezis works.
The problem is, I wanted to find the solution explicitly and verify by hand it belongs to $H^{2}(mathbb{R}^3)$, given that the domain is so simple, not using a known theorem. The solution of the equation is given by:
$$u(x)= left( frac{e^{-|y|}}{4 pi |y|} ast f right)(x)$$
But when I try to find a weak second order derivative of $u$ using integration by parts (and I do so by splitting the integration domain in and inside a ball of radius $epsilon$), I get this very annoying term that doesn't allow me to conclude:
$$int_{B(0, epsilon)^c} partial_{x_1}^2 left(frac{1}{4 pi |x|} right) e^{-|x|} phi(x+y) dx=int_{B(0, epsilon)^c} left(frac{3x_1 ^2 - |x|^2}{4 pi |x|^5} right) e^{-|x|} phi(x+y) dx$$
Where $phi in C^{infty}_0 (mathbb{R}^3)$. Now, I understand that something like this was forced to appear. Indeed, I know that $Delta (1/|x|)= delta(x)$, so this isn't a surprise. However, that expression over there does not converge when $epsilon to 0$ (or t least I think so, there should be a singularity like $1/rho$ if you pass to spherical coordinates) and I would like to obtain a weak derivative for $u$ in which I put all the derivatives on the term $e^{-|x|}/(4 pi |x|)$ of the convolution.
EDIT: Turns out I can just cut-off a ball of radius $epsilon$ with a smooth compact supported radial function and then integrating over the angular part first I get either 0 or a constant I don’t remember. Gotta think why I can pass to the limit, but the pointeise convergence is straightforward, and the dominance should follow easiliy, somehow.
pde sobolev-spaces distribution-theory calculus-of-variations regularity-theory-of-pdes
asked Dec 1 '18 at 13:58
tommy1996q
586414
add a comment |
In Brezis's book "Functional Anlaysis" it is proven that the solutions of the Helmotz equation $u - Delta u=f$ where $f in L^2 (mathbb{Omega})$ belong to $H^2 (mathbb{Omega}) cap H^1 _0 (Omega)$, when $Omega$ is a bounded domain of $mathbb{R}^n$ with $partial Omega$ regular enough. I think he also takes the Dirichlet boundary conditions equal to 0, but the problem I am dealing with takes into account $Omega = mathbb{R}^n$, so it shouldn't be a problem to integrate by parts using sufficently large balls. The thing is, he proves the existence of such a solution by shifting the problem to a minum of a particular functional. He then proves this minumum (that is a weak solution of the equation) belongs to $H^{1}_0 (Omega)$. Then, later on in the book, he proves that under the hypothesis of some regularity for $partial Omega$, the solution is in $H^2(Omega)cap H^1 _0 (Omega)$, and he does that using the so called "method of Nirmberg's translations".
The fact is: I am dealing with the case $Omega= mathbb{R}^3$. I think the same results that were true for regular bounded domains should also be true in this case, I think it should be enough to consider the limit of domains $Omega_N = B(0,N)$ where the proof of Brezis works.
The problem is, I wanted to find the solution explicitly and verify by hand it belongs to $H^{2}(mathbb{R}^3)$, given that the domain is so simple, not using a known theorem. The solution of the equation is given by:
$$u(x)= left( frac{e^{-|y|}}{4 pi |y|} ast f right)(x)$$
But when I try to find a weak second order derivative of $u$ using integration by parts (and I do so by splitting the integration domain in and inside a ball of radius $epsilon$), I get this very annoying term that doesn't allow me to conclude:
$$int_{B(0, epsilon)^c} partial_{x_1}^2 left(frac{1}{4 pi |x|} right) e^{-|x|} phi(x+y) dx=int_{B(0, epsilon)^c} left(frac{3x_1 ^2 - |x|^2}{4 pi |x|^5} right) e^{-|x|} phi(x+y) dx$$
Where $phi in C^{infty}_0 (mathbb{R}^3)$. Now, I understand that something like this was forced to appear. Indeed, I know that $Delta (1/|x|)= delta(x)$, so this isn't a surprise. However, that expression over there does not converge when $epsilon to 0$ (or t least I think so, there should be a singularity like $1/rho$ if you pass to spherical coordinates) and I would like to obtain a weak derivative for $u$ in which I put all the derivatives on the term $e^{-|x|}/(4 pi |x|)$ of the convolution.
EDIT: Turns out I can just cut-off a ball of radius $epsilon$ with a smooth compact supported radial function and then integrating over the angular part first I get either 0 or a constant I don’t remember. Gotta think why I can pass to the limit, but the pointeise convergence is straightforward, and the dominance should follow easiliy, somehow.
pde sobolev-spaces distribution-theory calculus-of-variations regularity-theory-of-pdes
asked Dec 1 '18 at 13:58
tommy1996q
586414
Why wouldn't it converge? The $exp(-lvert xrvert)$ is there to help for large $lvert xrvert$.
– user10354138
Dec 1 '18 at 14:17
I mean convergence for $epsilon to 0$. There I have a singularity of order $1/ rho$, passing in spherical coordinates. I'll edit the question, though.
– tommy1996q
Dec 1 '18 at 14:26
add a comment |
In Brezis's book "Functional Anlaysis" it is proven that the solutions of the Helmotz equation $u - Delta u=f$ where $f in L^2 (mathbb{Omega})$ belong to $H^2 (mathbb{Omega}) cap H^1 _0 (Omega)$, when $Omega$ is a bounded domain of $mathbb{R}^n$ with $partial Omega$ regular enough. I think he also takes the Dirichlet boundary conditions equal to 0, but the problem I am dealing with takes into account $Omega = mathbb{R}^n$, so it shouldn't be a problem to integrate by parts using sufficently large balls. The thing is, he proves the existence of such a solution by shifting the problem to a minum of a particular functional. He then proves this minumum (that is a weak solution of the equation) belongs to $H^{1}_0 (Omega)$. Then, later on in the book, he proves that under the hypothesis of some regularity for $partial Omega$, the solution is in $H^2(Omega)cap H^1 _0 (Omega)$, and he does that using the so called "method of Nirmberg's translations".
The fact is: I am dealing with the case $Omega= mathbb{R}^3$. I think the same results that were true for regular bounded domains should also be true in this case, I think it should be enough to consider the limit of domains $Omega_N = B(0,N)$ where the proof of Brezis works.
The problem is, I wanted to find the solution explicitly and verify by hand it belongs to $H^{2}(mathbb{R}^3)$, given that the domain is so simple, not using a known theorem. The solution of the equation is given by:
$$u(x)= left( frac{e^{-|y|}}{4 pi |y|} ast f right)(x)$$
But when I try to find a weak second order derivative of $u$ using integration by parts (and I do so by splitting the integration domain in and inside a ball of radius $epsilon$), I get this very annoying term that doesn't allow me to conclude:
$$int_{B(0, epsilon)^c} partial_{x_1}^2 left(frac{1}{4 pi |x|} right) e^{-|x|} phi(x+y) dx=int_{B(0, epsilon)^c} left(frac{3x_1 ^2 - |x|^2}{4 pi |x|^5} right) e^{-|x|} phi(x+y) dx$$
Where $phi in C^{infty}_0 (mathbb{R}^3)$. Now, I understand that something like this was forced to appear. Indeed, I know that $Delta (1/|x|)= delta(x)$, so this isn't a surprise. However, that expression over there does not converge when $epsilon to 0$ (or t least I think so, there should be a singularity like $1/rho$ if you pass to spherical coordinates) and I would like to obtain a weak derivative for $u$ in which I put all the derivatives on the term $e^{-|x|}/(4 pi |x|)$ of the convolution.
EDIT: Turns out I can just cut-off a ball of radius $epsilon$ with a smooth compact supported radial function and then integrating over the angular part first I get either 0 or a constant I don’t remember. Gotta think why I can pass to the limit, but the pointeise convergence is straightforward, and the dominance should follow easiliy, somehow.
pde sobolev-spaces distribution-theory calculus-of-variations regularity-theory-of-pdes
asked Dec 1 '18 at 13:58
tommy1996q
586414
In Brezis's book "Functional Anlaysis" it is proven that the solutions of the Helmotz equation $u - Delta u=f$ where $f in L^2 (mathbb{Omega})$ belong to $H^2 (mathbb{Omega}) cap H^1 _0 (Omega)$, when $Omega$ is a bounded domain of $mathbb{R}^n$ with $partial Omega$ regular enough. I think he also takes the Dirichlet boundary conditions equal to 0, but the problem I am dealing with takes into account $Omega = mathbb{R}^n$, so it shouldn't be a problem to integrate by parts using sufficently large balls. The thing is, he proves the existence of such a solution by shifting the problem to a minum of a particular functional. He then proves this minumum (that is a weak solution of the equation) belongs to $H^{1}_0 (Omega)$. Then, later on in the book, he proves that under the hypothesis of some regularity for $partial Omega$, the solution is in $H^2(Omega)cap H^1 _0 (Omega)$, and he does that using the so called "method of Nirmberg's translations".
The fact is: I am dealing with the case $Omega= mathbb{R}^3$. I think the same results that were true for regular bounded domains should also be true in this case, I think it should be enough to consider the limit of domains $Omega_N = B(0,N)$ where the proof of Brezis works.
The problem is, I wanted to find the solution explicitly and verify by hand it belongs to $H^{2}(mathbb{R}^3)$, given that the domain is so simple, not using a known theorem. The solution of the equation is given by:
$$u(x)= left( frac{e^{-|y|}}{4 pi |y|} ast f right)(x)$$
But when I try to find a weak second order derivative of $u$ using integration by parts (and I do so by splitting the integration domain in and inside a ball of radius $epsilon$), I get this very annoying term that doesn't allow me to conclude:
$$int_{B(0, epsilon)^c} partial_{x_1}^2 left(frac{1}{4 pi |x|} right) e^{-|x|} phi(x+y) dx=int_{B(0, epsilon)^c} left(frac{3x_1 ^2 - |x|^2}{4 pi |x|^5} right) e^{-|x|} phi(x+y) dx$$
Where $phi in C^{infty}_0 (mathbb{R}^3)$. Now, I understand that something like this was forced to appear. Indeed, I know that $Delta (1/|x|)= delta(x)$, so this isn't a surprise. However, that expression over there does not converge when $epsilon to 0$ (or t least I think so, there should be a singularity like $1/rho$ if you pass to spherical coordinates) and I would like to obtain a weak derivative for $u$ in which I put all the derivatives on the term $e^{-|x|}/(4 pi |x|)$ of the convolution.
EDIT: Turns out I can just cut-off a ball of radius $epsilon$ with a smooth compact supported radial function and then integrating over the angular part first I get either 0 or a constant I don’t remember. Gotta think why I can pass to the limit, but the pointeise convergence is straightforward, and the dominance should follow easiliy, somehow.
pde sobolev-spaces distribution-theory calculus-of-variations regularity-theory-of-pdes
pde sobolev-spaces distribution-theory calculus-of-variations regularity-theory-of-pdes
asked Dec 1 '18 at 13:58
tommy1996q
586414
asked Dec 1 '18 at 13:58
tommy1996q
586414
edited Dec 3 '18 at 18:24
asked Dec 1 '18 at 13:58
tommy1996q
586414
asked Dec 1 '18 at 13:58
tommy1996q
586414
asked Dec 1 '18 at 13:58
tommy1996q
586414
586414
Why wouldn't it converge? The $exp(-lvert xrvert)$ is there to help for large $lvert xrvert$.
– user10354138
Dec 1 '18 at 14:17
I mean convergence for $epsilon to 0$. There I have a singularity of order $1/ rho$, passing in spherical coordinates. I'll edit the question, though.
– tommy1996q
Dec 1 '18 at 14:26
add a comment |
Why wouldn't it converge? The $exp(-lvert xrvert)$ is there to help for large $lvert xrvert$.
– user10354138
Dec 1 '18 at 14:17
I mean convergence for $epsilon to 0$. There I have a singularity of order $1/ rho$, passing in spherical coordinates. I'll edit the question, though.
– tommy1996q
Dec 1 '18 at 14:26
Why wouldn't it converge? The $exp(-lvert xrvert)$ is there to help for large $lvert xrvert$.
– user10354138
Dec 1 '18 at 14:17
Why wouldn't it converge? The $exp(-lvert xrvert)$ is there to help for large $lvert xrvert$.
– user10354138
Dec 1 '18 at 14:17
I mean convergence for $epsilon to 0$. There I have a singularity of order $1/ rho$, passing in spherical coordinates. I'll edit the question, though.
– tommy1996q
Dec 1 '18 at 14:26
I mean convergence for $epsilon to 0$. There I have a singularity of order $1/ rho$, passing in spherical coordinates. I'll edit the question, though.
– tommy1996q
Dec 1 '18 at 14:26
add a comment |
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Why wouldn't it converge? The $exp(-lvert xrvert)$ is there to help for large $lvert xrvert$.
– user10354138
Dec 1 '18 at 14:17
I mean convergence for $epsilon to 0$. There I have a singularity of order $1/ rho$, passing in spherical coordinates. I'll edit the question, though.
– tommy1996q
Dec 1 '18 at 14:26